course Phy 202 005. `query 5
.............................................
Given Solution: ** The net force on the plug is P * A, where A is its cross-sectional area and P the pressure difference. • If L is the length of the plug then the net force F_net = P * A acts thru distance L doing work `dW = F_net * L = P * A * L. If the initial velocity of the plug is 0 and there are no dissipative forces, this is the kinetic energy attained by the plug. The volume of the plug is A * L so its mass is rho * A * L. • Thus we have mass rho * A * L with KE equal to P * A * L. Setting .5 m v^2 = KE we have .5 rho A L v^2 = P A L so that v = sqrt( 2 P / rho). Your Self-Critique: I solved the question like the problem set in which the mass of water was found using density (volume) and volume was found using cross-sectional area(height). I am not sure if my equation is the same as yours, just not using rho so could you check this and explain. Thanks.
.............................................
Given Solution: The density of air at common atmospheric temperature and pressure it about 1.3 kg / m^3. The volume of the room is 4.8 m * 3.8 m * 2.8 m = 51 m^3, approximately. The mass of the air in this room is therefore • mass = density * volume = 1.3 kg / m^3 * 51 m^3 = 66 kg, approximately. This is a medium-sized room, and the mass of the air in that room is close to the mass of an average-sized person. Your Self-Critique:OK Your Self-Critique Rating: OK ********************************************* Question: prin phy problem 10.8. Difference in blood pressure between head and feet of 1.60 m tell person. Pressure=rho *g *y 1000 kg / m^3 *9.8m/s^2 * 1.6m=1600N/m^2 1000kg/m^3 *9.8 m/s^2 * 0m = 0 N/m^2 The blood pressure differs by 1600N/m^2 ********************************************* Your Solution: Pressure=rho *g *y 1000 kg / m^3 *9.8m/s^2 * 1.6m=16000N/m^2 1000kg/m^3 *9.8 m/s^2 * 0m = 0 N/m^2 The blood pressure differs by about 16000N/m^2 Confidence Rating: 3
.............................................
Given Solution: The density of blood is close to that of water, 1000 kg / m^3. So the pressure difference corresponding to a height difference of 1.6 m is • pressure difference = rho g h = 1000 kg/m^3 * 9.80 m/s^2 * 1.60 m = 15600 kg / ( m s^2) = 15600 (kg m / s^2) / m^2 = 15600 N / m^2, or 15600 Pascals. 1 mm of mercury is 133 Pascals, so • 15,600 Pa = 15,600 Pa * ( 1 mm of Hg / 133 Pa) = 117 mm of mercury. Blood pressures are measured in mm of mercury; e.g. a blood pressure of 120 / 70 stands for a systolic pressure of 120 mm of mercury and a diastolic pressure of 70 mm of mercury. The density of blood is actually a bit higher than that of water; if you use a more accurate value for the density of blood you will therefore get a slightly great result. Your Self-Critique: I did not convert my findings to mmHg. It would be good to remember the conversion factor of 1N/m^2=1Pa and 1mm Hg= 133pascals Your Self-Critique Rating:3 ********************************************* Question: prin phy and gen phy 10.25 spherical balloon rad 7.35 m total mass 930 kg, Helium => what buoyant force ********************************************* Your Solution: R=7.35 Mass=930kg First we need to find the total volume of the balloon, the volume of a sphere is found using the equation V=(4/3) pi r^3 Which would give us V=1663.2m^3 We can then use volume to find mass as mass=density(v) And airs density is about 1.2 kg/m^3. This gives us a mass of 1995.8 kg. This is the mass of the air that was displaced by the balloon not the actually balloon. We then need to apply gravity’s force onto the balloon which would be 9.8m/s^2(1995.8 kg). This gives us 19558.8 N as the buoyant force. Confidence Rating: 1
.............................................
Given Solution: ** The volume of the balloon is about 4/3 pi r^3 = 1660 cubic meters and mass of air displaced is about 1.3 kg / m^3 * 1660 m^3 = 2160 kg. The buoyant force is equal in magnitude to the force of gravity on the displaced air, or about 2160 kg * 9.8 m/s^2 = 20500 Newtons, approx.. If the total mass of the balloon, including helium, is 930 kg then the net force is about • Net force = buoyant force - weight = 20,500 N - 9100 N = 11,400 N If the 930 kg doesn't include the helium we have to account also for the force of gravity on its mass. At about .18 kg/m^3 the 1660 m^3 of helium will have mass about 300 kg on which gravity exerts an approximate force of 2900 N, so the net force on the balloon would be around 11,400 N - 2900 N = 8500 N approx. The mass that can be supported by this force is m = F / g = 8500 N / (9.8 m/s^2) = 870 kg, approx.. ** Your Self-Critique: The solution solves for net force also. This is done by using the 930 kg and solving for the force gravity has on that giving about 9100 N. We can subtract this from the buoyant force to get a net force of about 10459N. The solution then goes on to determine how much helium would be supported by the net force of the balloon, I did not realize this was part of the problem. Your Self-Critique Rating: 3